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mathmath333

  • one year ago

Logarithm question

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align}& \text{solve}\hspace{.33em}\\~\\& \log_{2}(9-2^{x})=10^{\log_{10}(3-x)}\hspace{.33em}\\~\\ \end{align}}\)

  2. mathmath333
    • one year ago
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    @freegirl11112

  3. mathmath333
    • one year ago
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    @freckles

  4. anonymous
    • one year ago
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    hi I dot know this yet sorry

  5. mathmath333
    • one year ago
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    mistyped ur name was calling frekels

  6. freckles
    • one year ago
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    \[\log_2(9-2^x)=10^{\log(3-x)} \\ \log_2(9-2^x)=3-x \\ 2^{3-x}=9-2^x \\ 2^{3-x}+2^x=9 \\ 2^3 2^{-x}+2^x=9 \\ 8 \cdot 2^{-x}+2^x=9 \\ \text{ multiply both sides by } 2^x \\ 8 +2^{2x}=9 \cdot 2^{x}\]

  7. mathmath333
    • one year ago
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    i didn't understand the second step \(\log_2(9-2^x)=3-x\)

  8. Nnesha
    • one year ago
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    what happened to 10^log ?? :o

  9. freckles
    • one year ago
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    sorry log_10(x) and 10^x are inverses

  10. freckles
    • one year ago
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    \[10^{\log_{10}(3-x)}=3-x\] when 3-x>0

  11. Nnesha
    • one year ago
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    mhm oh okay thanks:=)

  12. freckles
    • one year ago
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    and I'm so sorry but I have to go someone is waiting on me

  13. freckles
    • one year ago
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    hey @mathmath333 were you able to finish

  14. freckles
    • one year ago
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    you have a quadratic in terms of 2^x

  15. mathmath333
    • one year ago
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    yep o got it.

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